Use $\epsilon-\delta$ language to show that $f(x)=x^\frac{1}{3}$ is differentiable at each point $x_0\neq0$ and not differentiable at $0$.
I've stared at this problem for a while and can't seem to figure out how to find $\delta$. How would I go about doing this?
edit: So I know that I want to find $\delta$ such that if $0<\mid x-x_0\mid<\delta$, then $\mid\frac{x^{1/3}-x_0^{1/3}}{x-x_0}-\frac{1}{3}x_0^{-2/3}\mid<\epsilon$. I can get to $$\mid\frac{x^{1/3}-x_0^{1/3}}{x-x_0}-\frac{1}{3}x_0^{-2/3}\mid=\mid\frac1{x^{2/3}+x^{1/3}x_0^{1/3}+x_0^{2/3}}-\frac{1}{3}x_0^{-2/3}\mid =\\ \mid\frac{1-\frac{1}{3}x^{2/3}x_0^{-2/3}-\frac{1}{3}x^{1/3}x_0^{-1/3}-\frac{1}{3}}{x^{2/3}+x^{1/3}x_0^{1/3}+x_0^{2/3}}\mid=\mid\frac{\frac{2}{3}-\frac{1}{3}x^{2/3}x_0^{-2/3}-\frac{1}{3}x^{1/3}x_0^{-1/3}}{x^{2/3}+x^{1/3}x_0^{1/3}+x_0^{2/3}}\mid$$ but I'm not sure how to get $\delta$ from here.
HINT:
For $h\ne 0$, note that
$$\frac{(x_o+h)^{1/3}-x_0^{1/3}}h=\frac1{(x_0+h)^{2/3}+(x_0+h)^{1/3}x_0^{1/3}+x_0^{2/3}}$$
Start by requiring that $\delta> |x_0/2|>0$. Can you finish now?